Question

An electric bulb, an open coil inductor, an ac source and a key are all connected in series to form a closed circuit. The key is closed and after some time an iron rod is inserted into the interior of the inductor, then

• A. The glow of the bulb increases
• B. The glow of the bulb remains unchanged
• C. The glow of the bulb decreases
• D. The bulb does not glow

Hint:

Remember the role of core materials in inductors: inserting a soft iron core into an inductor greatly increases its inductance. In a DC circuit, an inductor acts as a short circuit after the initial transient, so inserting an iron core would have no effect on the steady-state brightness. However, in an AC circuit, its reactance is crucial.

Answer 1

The Correct Option is C

Step 1: Understanding the circuit. The circuit is an RL series AC circuit. The current \(I\) determines the brightness of the bulb. \[ I = \frac{V_{rms}}{Z} = \frac{V_{rms}}{\sqrt{R^2 + X_L^2}} \] where \(R\) is the resistance of the bulb and \(X_L = \omega L\) is the inductive reactance.
Step 2: Effect of inserting an iron rod. The self-inductance of a coil is given by \( L = \mu_r \mu_0 n^2 V \). Iron is a ferromagnetic material with high relative permeability (\(\mu_r \gg 1\)). When the iron rod is inserted, the effective permeability increases significantly, causing the inductance \(L\) to increase.
Step 3: Effect on Current. As \(L\) increases, the reactance \(X_L = \omega L\) increases. Consequently, the total impedance \(Z = \sqrt{R^2 + X_L^2}\) increases. Since impedance increases, the current \(I\) flowing through the circuit decreases. As the current decreases, the power dissipated by the bulb (\(P = I^2 R\)) decreases, reducing its glow.